plus(

plus(0,

plus(s(

times(0,

times(s(0),

times(s(

div(0,

div(

div(div(

quot(0, s(

quot(s(

quot(

R

↳Dependency Pair Analysis

PLUS(s(x),y) -> PLUS(x,y)

TIMES(s(x),y) -> PLUS(y, times(x,y))

TIMES(s(x),y) -> TIMES(x,y)

DIV(x,y) -> QUOT(x,y,y)

DIV(div(x,y),z) -> DIV(x, times(y,z))

DIV(div(x,y),z) -> TIMES(y,z)

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**PLUS(s( x), y) -> PLUS(x, y)**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

The following dependency pair can be strictly oriented:

PLUS(s(x),y) -> PLUS(x,y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(PLUS(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

**TIMES(s( x), y) -> TIMES(x, y)**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

The following dependency pair can be strictly oriented:

TIMES(s(x),y) -> TIMES(x,y)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(TIMES(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

**QUOT( x, 0, s(z)) -> DIV(x, s(z))**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y),z) -> QUOT(x,y,z)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(QUOT(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(DIV(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

**QUOT( x, 0, s(z)) -> DIV(x, s(z))**

plus(x, 0) ->x

plus(0,y) ->y

plus(s(x),y) -> s(plus(x,y))

times(0,y) -> 0

times(s(0),y) ->y

times(s(x),y) -> plus(y, times(x,y))

div(0,y) -> 0

div(x,y) -> quot(x,y,y)

div(div(x,y),z) -> div(x, times(y,z))

quot(0, s(y),z) -> 0

quot(s(x), s(y),z) -> quot(x,y,z)

quot(x, 0, s(z)) -> s(div(x, s(z)))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes